# Thread: Backwards intergration

1. ## Backwards intergration

$\int\!\frac{1}{(1+x^2)^2}dx$ where x = tan u and u = arctan x

I can't see where acrtan has come from. What part of the integrand is arctan?

2. ## Re: Backwards intergration

They chose x= tan(u) in order to use the trig identity $1+ tan^2(u)= sec^2(u)$. There is NO part of the integral that is arctan. You need u= arctan x when you want to convert the completed integral back to x.

The point is that if you let x= tan(u) then $1+ x^2= 1+ tan^2 u= sec^2(u)$ while $dx= sec^2(u)du$ so your integral becomes
$\int \frac{sec^2(u)du}{(sec^2(u))^2}= \int\frac{sec^2(u)du}{sec^4(u)}=\int \frac{du}{sec^2(u)}= \int cos^2(u)du$. You can integrate that with the trig identity
$cos^2(u)= (1/2)(1+ cos(2u))$.

3. ## Re: Backwards intergration

Excellent, thanks for the help.